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A hypothetical (and maybe practical) question has been nagging at me.

If you had a piece of paper with dimensions 4 and 3 (4:3), folding it in half along the long side (once) would result in 2 inches and 3 inches (2:3), which wouldn't retain its ratio. For example, here is a piece of paper that doesn't retain its ratio when folded: enter image description here

Is retaining the ratio technically possible? If so, what is the side length and ratio that fulfills this requirement? Any help would be appreciated.

Update:

I added "once" because I got an answer saying that any recectangle would work, as any rectangle folded twice has the original ratio. Nice answer, but not quite what I was looking for. As for the other answers, I got 3x as much information as I needed! Thanks!

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    $\begingroup$ This is exactly why the whole world use A series paper (except US, of course). $\endgroup$ – Derek 朕會功夫 Feb 12 '15 at 4:41
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    $\begingroup$ Well, I have watched a video on YouTube about this and A series paper does retain its ratio length to width when folded in half width to width and it retains the same ratio. BTW, you still made this question smokin' hot! $\endgroup$ – ReliableMathBoy Feb 14 '15 at 5:07
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    $\begingroup$ Yes. It solves a big problem: printing documents (or images) scaled up or down to different paper size. Good luck with US letter. BTW check out NumberHub video about paper sizes where it is explained quite good. $\endgroup$ – Hauleth Feb 15 '15 at 2:10
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    $\begingroup$ @LokiClock: Some of us are simply curious about the world in which we live. We do not require facing "a particular problem" in order to wonder about something. It's called... imagination! :) $\endgroup$ – Lightness Races with Monica Feb 16 '15 at 11:34
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    $\begingroup$ @LightnessRacesinOrbit Others of us are interested in the people about us and their thought processes and style. The A series of paper sizes were chosen so as to have this property because it solves particular problems, mainly that if you design a print for one scale and have to drastically resize it, your rescaled print won't also have to be reformatted - nothing will get cut off, and it will fill the same portion of the page. Since the OP spoke in terms of halving paper, it's possible they could have rediscovered one of those problems, or found a new one. An artist might, for instance. $\endgroup$ – Loki Clock Feb 17 '15 at 12:23
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The $1:\sqrt{2}$ ratio ensures exactly that. That is the idea behind the ISO 216 standard for paper sizes, which was adopted from the German DIN 476 standard.

Its most common usage is the A series which especially in Europe is a collection of very common paper sizes. The base size, A0, has an area of a square meter, and every next smaller paper size is constructed by folding it in half.

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    $\begingroup$ @Mehrdad: because you don't use it, or because you took the ability to fold A4 in half to create an A5 booklet totally for granted? ;-) $\endgroup$ – Steve Jessop Feb 11 '15 at 10:55
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    $\begingroup$ @SteveJessop: Neither -- I don't use it currently, but I've used it before, and I simply have never noticed this. It's not like when you fold a piece of A4 paper you suddenly see a glowing "A5" that tells you it's an A5 paper... I've never dealt with A5, so I never knew what its dimensions were and how they related to A4 and such. $\endgroup$ – Mehrdad Feb 11 '15 at 11:23
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    $\begingroup$ It also means you can print booklets (2/4 pages on 1 sheet) without ending up with loads of whitespace. $\endgroup$ – NickG Feb 11 '15 at 12:30
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    $\begingroup$ @Mehrdad Actually not only A4 has the property, but the whole A format sequence has it: the A0 format is defined as a sheet of area 1 square meter and sides in proportion $\sqrt 2 : 1$, then every next $A_{n+1}$ is a half of $A_n$, retaining $\sqrt 2 : 1$ sides' lengths ratio. $\endgroup$ – CiaPan Feb 11 '15 at 16:59
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    $\begingroup$ "Actually not only A4 has the property, but the whole A format sequence has it" So do the B, C (and D) series. Because their side ratio is 1:sqrt(2), which is the key, as the other answers mention. As this is the accepted answer, it should mention that fact. $\endgroup$ – David Balažic Feb 11 '15 at 17:48
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A ratio of $1:\sqrt{2}$ will do the trick!

The original rectangle will have a ratio of $x:y$, where $y$ is the longer side and $x$ the shorter side. Then the folded rectangle will have a ratio of $\frac{1}{2}y:x$ and we want

$$\begin{align} \frac{x}{y} & =\frac{\frac{1}{2}y}{x} \\ x^2 & = \tfrac{1}{2}y^2 \\ x & = \tfrac{1}{\sqrt{2}} \cdot y \\ y & = \sqrt{2}\cdot x \end{align}$$

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    $\begingroup$ That's the unique solution, which is something that you should stress in the answer. $\endgroup$ – David Heffernan Feb 11 '15 at 15:42
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Well, suppose you have a rectangle of sides with $A$ and $B$ of length, $A$ being the bigger side.

When we fold it along the longest side we end up with a rectangle with sides of length $B$ and $A/2$.

So what you want to know is if there are any values of A and B that satisfy the following condition:

$$\frac{A}{B} = \frac{B}{A/2}$$

From solving the equation we get that $A = B \cdot \sqrt2$. So if the paper's height is $\sqrt2$ times its width, then we can make a rectangle with the properties that you wanted. And this is the only ratio that will work.

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$H:W = W:(H/2)$ resolves to $H:W = \sqrt 2$

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Standard Bond paper sizes A4, A3, A2 and A1 have been designed so that areas double up and sides increase by scale $ \sqrt 2 $.

$$ \frac LB = \frac 12 \cdot \frac bl $$

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  • $\begingroup$ The descriptor "Standard Bond" has nothing to do with the size of the paper. $\endgroup$ – Lightness Races with Monica Feb 16 '15 at 11:36
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When you fold any rectangular sheet of paper through the middle, clearly the original area is the double of resulting (folded) area.

At the same time, if you have a little rectangle with sides $a$ and $b$ (so area $ab$) (let us say $a>b$), and you want a similar rectangle with area $2ab$, then clearly that larger rectangle must have sides $\sqrt{2}\cdot a$ and $\sqrt{2}\cdot b$, the factor of magnification being $\sqrt{2}$.

Combining these two simple observations, and looking at your illustrations, we see that the long side $a$ of the "half" rectangle is equal to the short side $\sqrt{2}\cdot b$ of the "unfolded" (full) rectangle, so $a = \sqrt{2}\cdot b \Rightarrow \frac{a}{b} = \sqrt{2}$.

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Well, if you have paper that has a ratio equivalent to the square root of two with the length divided by the width, if you fold it in half hamburger style (from width to width), you will get another ratio that's equivalent to the square root of two of the same property, likewise if you fold it more in half that way.

How It Works:

$$\sqrt{2}/1=\sqrt{2}$$$$\sqrt{2}/2=1/\sqrt{2}=0.7071...$$

That's half the square root of two, and when one is divided by it, we get the square root of two again: $1/0.7071...=\sqrt{2}$ and it just keeps going on and on from number to number being divided by the square root of two to get that result. Just divide the dividend by the quotient in those things.

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A degenerate rectangle with an aspect ratio of $0$, also known as a line segment, can be folded in half either along its length or across its middle, yielding either the same segment or a segment having half the original length. Either way, the aspect ratio remains $0$. Algebraically, the solutions to the equations $$\frac{a}{b}=\frac{a/2}{b}$$ and $$\frac{a}{b}=\frac{a}{b/2}$$ are both $a=0$. (In order to speak of the ratio being a real number, I'm ruling out $b=0$.)

As the other answers observe, the non-degenerate cases are $$\frac{a}{b}=\frac{b}{a/2},$$ which has the solution $a=b\sqrt2$, and $$\frac{a}{b}=\frac{b/2}{a},$$ which has the solution $a=b/\sqrt2$.

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protected by Emily Feb 16 '15 at 3:07

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